What is the vertices ((x-5)^2)/9+((y-3)^2)/(16)=1 ?

The vertices ((x-5)^2)/9+((y-3)^2)/(16)=1 is (5,7),(5,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x-5)^2)/9+((y-3)^2)/(16)=1

vertices

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

(5, 7), (5, - 1)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (5, 3), b = 4, a = 3

(5, 3 + 4), (5, 3 - 4)

Refine

(5, 7), (5, - 1)

x^{2} + 9 y^{2} = 36

\frac{x ^{2}}{121} + (\frac{y}{1})^{2} = 1

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

6 x^{2} + 9 y^{2} - 54 = 0

+ 36 x^{2} + y^{2} = 36

What is the vertices ((x-5)^2)/9+((y-3)^2)/(16)=1 ?
The vertices ((x-5)^2)/9+((y-3)^2)/(16)=1 is (5,7),(5,-1)